- #1

Sandbo

- 18

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I have been having this question for a long while.

From Pozar's book (Microwave Engineering), on page 336, when he was solving for the resonance response and the Q factor for a Lamba/2 resonator terminated by an open circuit, and coupled to the feed line through a series capacitor, at some points he mentioned:

"...an uncoupled lambda/2 open-circuited transmission line resonator looks like a parallel RLC circuit near resonance, however, in the present case, the series coupling capacitor has the effect of inverting the driving point impedance, thus the response is now like a series RLC resonator..."

I didn't really understand how this happens with a series capacitor.

My understanding was such that, with having a series capacitor, I should take the points where the input impedance being zero (or lowest) as the resonance point, as frequency goes up the series coupling capacitor will any way make the input impedance infinite. (thus it doesn't make sense to consider a parallel RLC circuit which resonates with infinite (or highest) input impedance).

These days I was considering a series capacitively coupled short-circuited transmission line resonator (differs only from the above by the terminator)

What puzzled me further was, when I was simulating the above in HFSS, and plotting the Magnitude of the Input Impedance, I can see it going down slightly then going up a lot to form a peak. That way it in fact looks more like a parallel RLC circuit to me.

How should I perceive the concept of series coupling capacitor as impedance inverter?

And to apply it to various transmission line resonator?

Thanks a lot for your kind attention and my apology for the long post.

Any response would be much appreciated.